Answer the following questions using the standard potential, E°, values given in the table. Half rea — Analytical Chemistry Chemistry Question
Equilibrium constant
Answer the following questions using the standard potential, E°, values given in the table.
Half reaction E°/ V (298 K)
Sn2+ + 2 e– → Sn –0.14
Sn4+ + 2 e– → Sn2+ +0.15
Hg2 2+ + 2 e– → 2 Hg +0.79
Hg2Cl2 + 2 e– → 2 Hg + 2 Cl– +0.27
Calculate the equilibrium constant, K, for the following reaction at 298 K
Sn(s) + Sn4+(aq) ⇌ 2 Sn2+(aq)
Model Answer
R: Sn4+ + 2 e– → Sn2+ + 0.15 V
L: Sn2+ + 2 e– → Sn – 0.14 V
E° = + 0.29 V
K = b_e(Sn4+) / b_e(Sn2+)^2 = exp(22.59) = 6.4×10^9.
Calculate the solubility, S, of Hg2Cl2 in water at 298 K (units for S, mol kg–1). The mercury cation in the aqueous phase is Hg2 2+.
Model Answer
R: Hg2Cl2 + 2 e– → 2 Hg + 2 Cl– + 0.27
L: Hg2 2+ + 2 e– → 2 Hg + 0.79
E° = –0.52 V
K = be(Hg2 2+) ^2 = exp(– 40.50) = 2.58×10–18
let x = be(Hg2 2+), then be(Cl–) = 2x. Therefore, K = x(2x)^2 = 4x^3
Solving this equation for x gives: x = (K / 4)^(1/3) = 8.6×10–7
S = x = 8.6×10–7 mol kg–1.
Calculate the voltage, E°, of a fuel cell by using the following reaction involving two electrons.
H2(g) + 1/2 O2(g) → H2O(l) ∆rG° = –237.1 kJ mol–1
Model Answer
E° = –∆rG° / nF = –(–237.1 × 1000) / (2 × F) = 1.23 V